(A) The Nernst equation for the hydrogen electrode reaction is given by:$E_{H^+/H_2} = E^{\circ}_{H^+/H_2} - \frac{2.303 RT}{nF} \log \frac{P_{H_2}^{1

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(A) The Nernst equation for the hydrogen electrode reaction is given by:$E_{H^+/H_2} = E^{\circ}_{H^+/H_2} - \frac{2.303 RT}{nF} \log \frac{P_{H_2}^{1/2}}{[H^+]}$Given $E^{\circ}_{H^+/H_2} = 0 \ V$,$n = 1$,$P_{H_2} = 1 \ bar$,and $\frac{2.303 RT}{F} = 0.06 \ V$.Substituting these values:$E = 0 - 0.06 \log \frac{1}{[H^+]}$Since $pH = -\log[H^+]$,we have $\log \frac{1}{[H^+]} = pH = 10.0$.Therefore,$E = -0.06 \times 10.0 = -0.6 \ V$.

Class 12 Chemistry Electrochemistry Nernst equation and ECS

Medium

At $298 \ K$,the following reaction takes place for a hydrogen electrode:
$H^{+}_{(aq)} + e^{-} \longrightarrow \frac{1}{2} H_2(1 \ bar)$
The solution $pH$ is $10.0$. What is the hydrogen electrode potential in volts?
$\left(\frac{2.303 RT}{F} = 0.06 \ V\right)$

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A
$-0.6 \ V$
B
$-0.06 \ V$
C
$+0.6 \ V$
D
$+0.06 \ V$

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Nernst equation and ECS

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The change in electrode potential of $Cr^{3+}/Cr_2O_7^{2-}$ electrode at $25 \ ^oC$ due to change in $pH$ of its electrolytic solution from $1$ to $3$ is ........... $V$ (Assume: $[Cr_2O_7^{2-}]$ and $[Cr^{3+}]$ remain constant). Use: $\frac{2.303RT}{F} = 0.06$.

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At $298 \ K$,find out the $emf$ for the cell:
$Al_{(s)} | Al^{+3} (0.1 \ M) || Fe^{+2} (0.001 \ M) | Fe_{(s)}$
Given: $E^o_{Al^{+3}/Al} = -1.66 \ V$ and $E^o_{Fe^{+2}/Fe} = -0.44 \ V$.

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What will be the $pH$ of a solution which is in contact with a hydrogen electrode having an oxidation potential of $0.177 \ V$?

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Given $E^o _{Cr^{3+} / Cr} = -0.72 \, V$ and $E^o _{Fe^{2+} / Fe} = -0.42 \, V$. The potential for the cell $Cr \, | \, Cr^{3+}_{(0.1 \, M)} \, || \, Fe^{2+}_{(0.01 \, M)} \, | \, Fe$ is ......... $V$.

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If $E_{cell} = 0.118 \, V$ for the following reaction,calculate $[H^{+}]$ and $pH$ at $298 \, K$ temperature.
$Pt \mid H_2(1 \, bar) \mid H^{+} (10^{-6} \, M) \parallel H^{+} (x \, M) \mid H_2 (1 \, bar) \mid Pt$

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At $298 \ K$,the following reaction takes place for a hydrogen electrode:$H^{+}_{(aq)} + e^{-} \longrightarrow \frac{1}{2} H_2(1 \ bar)$The solution $pH$ is $10.0$. What is the hydrogen electrode potential in volts?$\left(\frac{2.303 RT}{F} = 0.06 \ V\right)$

Similar Questions

The change in electrode potential of $Cr^{3+}/Cr_2O_7^{2-}$ electrode at $25 \ ^oC$ due to change in $pH$ of its electrolytic solution from $1$ to $3$ is ........... $V$ (Assume: $[Cr_2O_7^{2-}]$ and $[Cr^{3+}]$ remain constant). Use: $\frac{2.303RT}{F} = 0.06$.

At $298 \ K$,find out the $emf$ for the cell:$Al_{(s)} | Al^{+3} (0.1 \ M) || Fe^{+2} (0.001 \ M) | Fe_{(s)}$Given: $E^o_{Al^{+3}/Al} = -1.66 \ V$ and $E^o_{Fe^{+2}/Fe} = -0.44 \ V$.

What will be the $pH$ of a solution which is in contact with a hydrogen electrode having an oxidation potential of $0.177 \ V$?

Given $E^o _{Cr^{3+} / Cr} = -0.72 \, V$ and $E^o _{Fe^{2+} / Fe} = -0.42 \, V$. The potential for the cell $Cr \, | \, Cr^{3+}_{(0.1 \, M)} \, || \, Fe^{2+}_{(0.01 \, M)} \, | \, Fe$ is ......... $V$.

If $E_{cell} = 0.118 \, V$ for the following reaction,calculate $[H^{+}]$ and $pH$ at $298 \, K$ temperature.$Pt \mid H_2(1 \, bar) \mid H^{+} (10^{-6} \, M) \parallel H^{+} (x \, M) \mid H_2 (1 \, bar) \mid Pt$

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At $298 \ K$,the following reaction takes place for a hydrogen electrode:
$H^{+}_{(aq)} + e^{-} \longrightarrow \frac{1}{2} H_2(1 \ bar)$
The solution $pH$ is $10.0$. What is the hydrogen electrode potential in volts?
$\left(\frac{2.303 RT}{F} = 0.06 \ V\right)$

At $298 \ K$,find out the $emf$ for the cell:
$Al_{(s)} | Al^{+3} (0.1 \ M) || Fe^{+2} (0.001 \ M) | Fe_{(s)}$
Given: $E^o_{Al^{+3}/Al} = -1.66 \ V$ and $E^o_{Fe^{+2}/Fe} = -0.44 \ V$.

If $E_{cell} = 0.118 \, V$ for the following reaction,calculate $[H^{+}]$ and $pH$ at $298 \, K$ temperature.
$Pt \mid H_2(1 \, bar) \mid H^{+} (10^{-6} \, M) \parallel H^{+} (x \, M) \mid H_2 (1 \, bar) \mid Pt$